Question

5 moles of an ideal gas (Cp = 3R, Cv 2R ) are heated from 25°C to 300°C. Calculate (a) The change in internal energy of the gas (b) The change in enthalpy of the gas.

Answer 1

(a) The change in internal energy of the gas is 22.86 kJ.

(b) The change in enthalpy of the gas is 34.29 kJ.

Explanation :

(a) The formula used for change in internal energy of the gas is:

`\Delta U=nC_v\Delta T\\\\\Delta U=nC_v(T_2-T_1)`

where,

`\Delta U` = change in internal energy = ?

n = number of moles of gas = 5 moles

`C_v` = heat capacity at constant volume = 2R

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature =
`25^oC=273+25=298K`

`T_2` = final temperature =
`300^oC=273+300=573K`

Now put all the given values in the above formula, we get:

`\Delta U=nC_v(T_2-T_1)`

`\Delta U=(5moles)* (2R)* (573-298)`

`\Delta U=(5moles)* 2(8.314J/mole.K)* (573-298)`

`\Delta U=22863.5J=22.86kJ`

The change in internal energy of the gas is 22.86 kJ.

(b) The formula used for change in enthalpy of the gas is:

`\Delta H=nC_p\Delta T\\\\\Delta H=nC_p(T_2-T_1)`

where,

`\Delta H` = change in enthalpy = ?

n = number of moles of gas = 5 moles

`C_p` = heat capacity at constant pressure = 3R

R = gas constant = 8.314 J/mole.K

`T_1` = initial temperature =
`25^oC=273+25=298K`

`T_2` = final temperature =
`300^oC=273+300=573K`

Now put all the given values in the above formula, we get:

`\Delta H=nC_p(T_2-T_1)`

`\Delta H=(5moles)* (3R)* (573-298)`

`\Delta H=(5moles)* 3(8.314J/mole.K)* (573-298)`

`\Delta H=34295.25J=34.29kJ`

The change in enthalpy of the gas is 34.29 kJ.